Let $f(x)=\dfrac{x^2-2}{x-1}$. $f'(x)=$
Explanation: $f$ is a rational function. To find the derivative of rational functions, we use the quotient rule : $\begin{aligned} \dfrac{d}{dx}\left[\dfrac{u(x)}{v(x)}\right]&=\dfrac{\dfrac{d}{dx}[u(x)]v(x)-u(x)\dfrac{d}{dx}[v(x)]}{[v(x)]^2} \\\\ &=\dfrac{u'(x)v(x)-u(x)v'(x)}{[v(x)]^2} \end{aligned}$ Let's differentiate! = f ′ ( x ) = d d x ( x 2 − 2 x − 1 ) = ( x − 1 ) d d x ( x 2 − 2 ) − ( x 2 − 2 ) d d x ( x − 1 ) ( x − 1 ) 2 = ( x − 1 ) ( 2 x ) − ( x 2 − 2 ) ( 1 ) ( x − 1 ) 2 = 2 x 2 − 2 x − x 2 + 2 ( x − 1 ) 2 = x 2 − 2 x + 2 ( x − 1 ) 2 The quotient rule Differentiate ( x 2 − 2 ) & ( x − 1 ) Expand \begin{aligned} &\phantom{=}f'(x) \\\\ &=\dfrac{d}{dx}\left(\dfrac{x^2-2}{x-1}\right) \\\\ &=\dfrac{(x-1)\dfrac{d}{dx}(x^2-2)-(x^2-2)\dfrac{d}{dx}(x-1)}{(x-1)^2}&&\gray{\text{The quotient rule}} \\\\ &=\dfrac{(x-1)(2x)-(x^2-2)(1)}{(x-1)^2}&&\gray{\text{Differentiate }(x^2-2)\text{ & }(x-1)} \\\\ &=\dfrac{2x^2-2x-x^2+2}{(x-1)^2}&&\gray{\text{Expand}} \\\\ &=\dfrac{x^2-2x+2}{(x-1)^2} \end{aligned} In conclusion, $f'(x)=\dfrac{x^2-2x+2}{(x-1)^2}$, or any other equivalent form.